Math, asked by aaravrathore18182g, 2 days ago

Find the PERIMETER of a right angled triangle with base 8 cm and hypotenuse. 17 cm.​

Answers

Answered by MystícαIStαr
69

Answer:

  • Perimeter of right angled traingle is 40cm

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Step-by-step explanation:

Here in the question we are given base and hypotenuse of the Traingle. So, firstly we find the perpendicular of the Traingle.

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Using Pythagoras theorem,

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 \sf \mapsto {(AC)}^{2}  =  {(AB)}^{2}  -  {(BC)}^{2}  \\  \\  \\  \sf \mapsto {(AC)}^{2}  =  {(17)}^{2} -  {(8)}^{2}   \\  \\  \\  \sf \mapsto {(AC)}^{2}  = 289  -  64  \\  \\  \\  \sf \mapsto {(AC)}^{2}  = 225 \\  \\  \\  \sf \mapsto AC =  \sqrt{225}  \\  \\  \\  \sf \mapsto \: \boxed{  \sf \pmb{AC = 15m}} \:  \pink \star

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We have get the required Third side of the traingle. We know that perimeter of Traingle is the sum of all sides.

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  • { \boxed{ \frak { \pmb{Perimeter  \: of  \: Traingle =  a + b + c}}}}

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 \sf \mapsto \ 17 + 8 + 15 \\  \\  \\  \sf \mapsto 17 + 23 \\  \\  \\  \sf \mapsto  {\pmb{{\boxed{ \rm{40 { \: cm} }}}}} \:  \pink \star

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  • Perimeter of right angled Traingle is 40cm.
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Answered by PopularStar
35

\text\orange{SolutioN:}

In right angle triangle:

CB=8 cm

AB=17 cm

AC=?

\sf\pink{(AC)^2}=\sf\pink{(AB)^2}-\sf\pink{(CB)^2}\text\orange{(By \ pythagoras \ theorem)}

\sf\pink{(AC)^2}=\sf\pink{(17)^2}-\sf\pink{(8)^2}

\sf\pink{(AC)^2}=\sf\pink{289}-\sf\pink{64}

\sf\pink{AC^2}=225

\sf\pink{AC}=15 cm

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Now,

Perimeter of right angle triangle=A+B+C

∴17+8+15=\text \green{40 cm}

Hence, Perimeter of right angle triangle is \sf \red{40 \ cm...}

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